What is log_{3}(2)+log_{3}(5)-log_{3}(4) ?

The solution to log_{3}(2)+log_{3}(5)-log_{3}(4) is log_{3}(5/2)

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simplify log_{3}(2)+log_{3}(5)-log_{3}(4)

simplify

lo g_{3} (2) + lo g_{3} (5) - lo g_{3} (4)

lo g_{3} (\frac{5}{2})

Decimal Notation

0.83404 \dots

Solution steps

lo g_{3} (2) + lo g_{3} (5) - lo g_{3} (4)

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{3} (2) + lo g_{3} (5) = lo g_{3} (2 \cdot 5)

= lo g_{3} (2 \cdot 5) - lo g_{3} (4)

Multiply the numbers:

2 \cdot 5 = 10

= lo g_{3} (10) - lo g_{3} (4)

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{3} (10) - lo g_{3} (4) = lo g_{3} (\frac{10}{4})

= lo g_{3} (\frac{10}{4})

Cancel the common factor:

2

= lo g_{3} (\frac{5}{2})

∣ 5 x + 2 ∣ - ∣ x^{2} + 2 ∣ + ln (4 x)

2 ln (x + 1) - ln (x)

1.03 ln (2.538 x)

ln (x - 4) [ln (x + 3) + ln (x - 5)]

\frac{1}{2} ln (x + 1) + \frac{1}{2} ln (x - 1)

What is log_{3}(2)+log_{3}(5)-log_{3}(4) ?
The solution to log_{3}(2)+log_{3}(5)-log_{3}(4) is log_{3}(5/2)